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Markdown+itex2MML Sandbox

Play around below. Your changes will, periodically, be rolled back.

Some examples

(1)minw hp h+w rp r+w lp l \mathop{min} w_h p_h + w_r p_r + w_l p_l

Here’s an equation

(2) e ax 2/2dx=2πa {\int_{-\infty}^\infty e^{-a x^2/2} \mathrm{d}x} = \sqrt{\frac{2\pi}{a}}

which we can later refer1 back to as (2).

Aligned equations:

(3)a+b =b+a a+(b+c) =(a+b)+c\begin{aligned} a+b &= b+a \\ a+(b+c) &= (a+b)+c \end{aligned}

The Dirac equation (boxed):

(iD+m)ψ=0 \boxed{(i\slash{D}+m)\psi = 0}

Here’s the table of Clifford2 algebras over \mathbb{R}:

jj001122334455667788
𝒞 j \mathcal{C}\ell_{j}^-\mathbb{R}\mathbb{C}\mathbb{H}\mathbb{H}\oplus\mathbb{H}(2)\mathbb{H}(2)(4)\mathbb{C}(4)(8)\mathbb{R}(8)(8)(8)\mathbb{R}(8)\oplus\mathbb{R}(8)(16)\mathbb{R}(16)
𝒞 j +\mathcal{C}\ell_{j}^+\mathbb{R}\mathbb{R}\oplus\mathbb{R}(2)\mathbb{R}(2)(2)\mathbb{C}(2)(2)\mathbb{H}(2)(2)(2)\mathbb{H}(2)\oplus\mathbb{H}(2)(4)\mathbb{H}(4)(8)\mathbb{C}(8)(16)\mathbb{R}(16)

where the generators of 𝒞 j ±\mathcal{C}\ell_{j}^\pm satisfy

γ iγ j+γ jγ i=±2δ ij \gamma_i\gamma_j +\gamma_j \gamma_i =\pm 2\delta_{i j}

and 𝒞 n+8 ±=𝒞 n ±(16)\mathcal{C}\ell_{n+8}^\pm = \mathcal{C}\ell_n^\pm \otimes \mathbb{R}(16).

(4)lim n1 k=1 n1k 2=π 26 \lim_{n \to \infty-1} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}

More Examples

(5)×E=Bt\nabla \times \vec{E} = - \frac {\partial \vec{B}}{\partial t}
(6)Bdl=μ 0I enc \oint \mathbf{B}\cdot \mathrm{d}\mathbf{l} = \mu_0 I_\text{enc}

H 1(𝒵,𝒪(k))H^1(\mathcal{Z}, \mathcal{O}(-k)) Let G=(V,E)G=(V,E) be a graph, with w:V[0,1]w:V\to [0,1] a weight function.

(7){Q i,Q j}=δ ij.\{Q_i, Q_j\} = \delta_{ij}\mathcal{H}.

Theorems

Definition

Let HH be a subgroup of a group GG. A left coset of HH in GG is a subset of GG that is of the form xHx H, where xGx \in G and xH={xh:hH}x H = \{ x h : h \in H \}.

Similarly a right coset of HH in GG is a subset of GG that is of the form HxH x, where Hx={hx:hH}H x = \{ h x : h \in H \}.

Lemma

Let HH be a subgroup of a group GG, and let xx and yy be elements of GG. Suppose that xHyHx H \cap y H is non-empty. Then xH=yHx H = y H.

Proof

Let zz be some element of xHyHx H \cap y H. Then z=xaz = x a for some aHa \in H, and z=ybz = y b for some bHb \in H. If hh is any element of HH then ahHa h \in H and a 1hHa^{-1}h \in H, since HH is a subgroup of GG. But zh=x(ah)z h = x(a h) and xh=z(a 1h)xh = z(a^{-1}h) for all hHh \in H. Therefore zHxHz H \subset x H and xHzHx H \subset z H, and thus xH=zHx H = z H. Similarly yH=zHy H = z H, and thus xH=yHx H = y H, as required.

Lemma

Let HH be a finite subgroup of a group GG. Then each left coset of HH in GG has the same number of elements as HH.

Proof

Let H={h 1,h 2,,h m}H = \{ h_1, h_2,\ldots, h_m\}, where h 1,h 2,,h mh_1, h_2,\ldots, h_m are distinct, and let xx be an element of GG. Then the left coset xHx H consists of the elements xh jx h_j for j=1,2,,mj = 1,2,\ldots,m. Suppose that jj and kk are integers between 11 and mm for which xh j=xh kx h_j = x h_k. Then h j=x 1(xh j)=x 1(xh k)=h kh_j = x^{-1} (x h_j) = x^{-1} (x h_k) = h_k, and thus j=kj = k, since h 1,h 2,,h mh_1, h_2,\ldots, h_m are distinct. It follows that the elements xh 1,xh 2,,xh mx h_1, x h_2,\ldots, x h_m are distinct. We conclude that the subgroup HH and the left coset xHx H both have mm elements, as required.

Theorem

(Lagrange’s Theorem). Let GG be a finite group, and let HH be a subgroup of GG. Then the order of HH divides the order of GG.

Proof

Each element xx of GG belongs to at least one left coset of HH in GG (namely the coset xHx H), and no element can belong to two distinct left cosets of HH in GG (see Lemma 1). Therefore every element of GG belongs to exactly one left coset of HH. Moreover each left coset of HH contains |H||H| elements (Lemma 2). Therefore |G|=n|H||G| = n |H|, where nn is the number of left cosets of HH in GG. The result follows.

Corollary

Let xx be an element of a finite group GG. Then the order of xx divides the order of GG.

Theorem

Let f:ΔΔ,f : \Delta \longrightarrow \Delta, where Δ={z:|z|<1}\Delta=\{z\in\mathbb{C}: \vert z \vert \lt 1\}, be analytic with aΔa \in \Delta. Then

|f(z)f(a)1f(a)¯f(z)||za1a¯z| \left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert\le \left\vert\frac{z-a}{1-\overline{a}z}\right\vert

for all |z|1\vert z \vert \le 1 and

|f(a)|1|f(a)| 211|a| 2. \frac{\vert f'(a)\vert}{1-\vert f(a)\vert^2}\le \frac{1}{1-\vert a \vert^2}.

Furthermore, equality holds iff ff realizes a conformal mapping of Δ\Delta onto itself.

Proof

Let w=za1a¯zw=\frac{z-a}{1-\overline{a}z} and put ϕ(w)=f(z)f(a)1f(a)¯f(z)\phi(w)=\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}. Define for absb<1\abs{b}\lt 1 C b(z)=zb1b¯z.C_b(z)=\frac{z-b}{1-\overline{b}z}. All conformal maps from Δ\Delta to itself, sending bb to 00, are of the form C b(z)e iγC_b(z)e^{i\gamma} for γ[0,2π].\gamma\in[0,2\pi]. In this notation, ϕ(w)=C f(a)fC a 1(w),\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w), where C a 1C_a^{-1} is the inverse of C aC_a as a function. Note that C a(z)C_a(z) is conformal, so it has an inverse. It is clear that ϕ(0)=C f(a)fC a 1(0)=C f(a)(f(a))=0\phi(0)=C_{f(a)}\circ f \circ C_a^{-1}(0)=C_{f(a)}(f(a))=0. Since C a 1:ΔΔC_a^{-1}: \Delta \longrightarrow \Delta and f:ΔΔf : \Delta \longrightarrow \Delta and C f(a):ΔΔ,C_{f(a)}: \Delta \longrightarrow \Delta, then |ϕ(w)|<1\vert\phi(w)\vert\lt 1 for |w|<1.\vert w\vert \lt 1 . Applying Schwarz’s lemma, we obtain |ϕ(w)||w|\vert\phi(w)\vert\le \vert w \vert for |w|1\vert w \vert \le 1. Furthermore, if equality holds, then f(z)=e iγzf(z)=e^{i\gamma'} z for γ[0,2π]\gamma'\in [0,2\pi]. Therefore,

(8)|f(z)f(a)1f(a)¯f(z)||za1a¯z| \left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert\le \left\vert\frac{z-a}{1-\overline{a}z}\right\vert

for all |z|1.\vert z \vert\le 1. Rearranging, we obtain

|f(z)f(a)za||1f(a)¯f(z)1a¯z|. \left\vert\frac{f(z)-f(a)}{z-a}\right\vert\le\left\vert\frac{1-\overline{f(a)}f(z)}{1-\overline{a}z}\right\vert.

If we take the limit as zz tends to aa, we obtain

|f(a)||1|f(a)| 21|a| 2|=1|f(a)| 21|a| 2, \left\vert f'(a)\right\vert \le \left\vert\frac{1-\vert f(a)\vert ^2}{1-\vert a \vert^2}\right\vert=\frac{1-\vert f(a)\vert^2}{1-\vert a \vert^2},

or

|f(a)|1|f(a)| 211|a| 2. \frac{\vert f'(a)\vert}{1-\vert f(a)\vert^2}\le \frac{1}{1-\vert a \vert^2}.

As said above, if equality holds in (8), then Schwarz’s lemma tells us that ϕ(w)=e iγw\phi(w)=e^{i\gamma'}w. Thus, ϕ(w)=C f(a)fC a 1(w)=e iγw,\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w)=e^{i\gamma'}w, so f(z)=C f(a) 1(e iγC a(z))f(z)=C_{f(a)}^{-1}(e^{i\gamma'}C_a(z)). Since e iγC a(z)e^{i\gamma'}C_a(z) is conformal, C f(a) 1,C_{f(a)}^{-1}, the inverse function of C f(a)C_{f(a)}, is conformal, and a composition of conformal maps is conformal, then ff is a conformal map of Δ\Delta onto itself. Conversely, if ff is a conformal map of Δ\Delta onto itself, then ϕ(w)=C f(a)fC a 1(w)=e iγC b(w),\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w)=e^{i\gamma}C_b(w), since a composition of conformal maps is conformal and because all conformal maps from Δ\Delta onto itself are of the form e iγC b(w).e^{i\gamma}C_b(w). We also know that ϕ(0)=0,\phi(0)=0, so b=0b=0. Therefore,

ϕ(w)=e iγC 0(w)=e iγw|ϕ(w)|=|w||f(z)f(a)1f(a)¯f(z)|=|za1a¯z| \phi(w)=e^{i\gamma}C_0(w)=e^{i\gamma}w \Leftrightarrow \vert\phi(w)\vert=\vert w \vert \Leftrightarrow \left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert=\left\vert\frac{z-a}{1-\overline{a}z}\right\vert

for all |z|1\vert z \vert\le 1. In sum, equality holds in (8) iff ff is a conformal map from Δ\Delta to itself.

Remark

Someone needs to code \abs, i.e. \left\vert # \right\vert. This is terribly annoying. Also, when using $$, why must one enter a line before and after.

SVG graphics, created in SVG-Edit (Instiki 0.19 feature):

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Commutative cube Layer 1 A 0 A_0 C 0 C_0 A 1 A_1 C 1 C_1 B 0 B_0 D 0 D_0 B 1 B_1 D 1 D_1
Box diagram W + W^+ W W^- s ¯ \overline{s} d ¯ \overline{d} Layer 1 s s d d u , c , t u,\, c,\, t u , c , t u,\, c,\, t


K 0K¯ 0K^0\overline{K}^0 Mixing

Yet More examples

  • SVG:
  • Animated SVG
Example anim01 - demonstrate animation elements e ax 2 d x = π a {\int_{-\infty}^{\infty}e^{-a x^2}d x}=\sqrt{\tfrac{\pi}{a}}
  • Complicated commutative diagrams (equations in SVG)
Complicated commutative diagram, realized in SVG 1 1 1 1 1 1 1 1 Id Id Id Id A A B B ρ \rho H H H H K K K ' K&#39; ϕ 1 \phi_1 ϕ 2 \phi_2 N A N_A N B N_B N A N^\vee_A N B N^\vee_B
  • SVG in equations.

In SU(3)SU(3), Rank-2 Symmetric Tensor Representation Fundamental Representation = Adjoint Representation Rank-3 Symmetric Tensor Representation \begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="30" height="16" viewBox="0 0 30 16"> <desc>Rank-2 Symmetric Tensor Representation</desc> <g transform="translate(5,5)" fill="#FCC" stroke="#000" stroke-width="2"> <rect width="10" height="10"/> <rect width="10" height="10" x="10"/> </g> </svg> \end{svg}\includegraphics[width=2em]{young1} \otimes \begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="20" height="16" viewBox="0 0 20 16"> <desc>Fundamental Representation</desc> <g transform="translate(5,5)" fill="#FCC" stroke="#000" stroke-width="2"> <rect width="10" height="10"/> </g> </svg> \end{svg}\includegraphics[width=1em]{young2} = \begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="30" height="26" viewBox="0 0 30 26"> <desc>Adjoint Representation</desc> <g transform="translate(5,5)" fill="#FCC" stroke="#000" stroke-width="2"> <rect width="10" height="10"/> <rect width="10" height="10" x="10"/> <rect width="10" height="10" y="10"/> </g> </svg> \end{svg}\includegraphics[width=2em]{young3} \oplus \begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="40" height="16" viewBox="0 0 40 16"> <desc>Rank-3 Symmetric Tensor Representation</desc> <g transform="translate(5,5)" fill="#FCC" stroke="#000" stroke-width="2"> <rect width="10" height="10"/> <rect width="10" height="10" x="10"/> <rect width="10" height="10" x="20"/> </g> </svg> \end{svg}\includegraphics[width=3em]{young4}.

  • Cases:
r a+1={0 with prob.exp(θr a) max{δr a,z} with prob.1exp(θr a) r_{a+1} = \begin{cases} 0 & \text{with prob.}\quad \exp(-\theta r_a) \\ \max \lbrace \delta r_a, z \rbrace & \text{with prob.}\quad 1 - \exp(-\theta r_a) \\ \end{cases}
  • Stretchy Brackets:
q a(z)=σ a 1exp[γ+zσ a]q_a(z) = \sigma_a^{-1} \exp{\left[ -\frac{\gamma + z}{\sigma_a} \right]}
  • Linearity of Quadrature Rules
i=1 N(αf(x i)+βg(x i))w i=α i=1 Nf(x i)w i+β i=1 Ng(x i)w i\sum_{i = 1}^N {\left( {\alpha f(x_i ) + \beta g(x_i )} \right)w_i } = \alpha \sum_{i = 1}^N {f(x_i )w_i } + \beta \sum_{i = 1}^N {g(x_i )w_i }
  • Linearity of Integrals
a b(αf(x)+βg(x))dx=α a bf(x)dx+β a bg(x)dx{\int_a^b {\left( {\alpha f(x)\, + \beta g(x)} \right)dx = } \alpha \int_a^b {f(x)\,dx} + \beta \int_a^b {g(x)\,dx} }
  • Can we talk about x i 2x_i^2 inline? What about a bx 2dx\int_a^b x^2\,dx? Inline fractions xx 2x 1x 2\frac{x-x_2}{x_1-x_2}?

  • Big fractions

p 3(x)=(12)(x12)(x34)(x1)(1412)(1434)(141)+(12)(x12)(x34)(x1)(1412)(1434)(141)+(12)(x12)(x34)(x1)(1412)(1434)(141)+(12)(x12)(x34)(x1)(1412)(1434)(141)p_3 (x) = \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}}
  • Diagram
P 1(Y) P 1(X) T T \begin{matrix} P_1(Y) &\to& P_1(X) \\ \downarrow &\Downarrow^\mathrlap{\sim}& \downarrow \\ T' &\to& T \end{matrix}

(9) A_n Quiver v1 v2 vn1 (U(k) n 1,{v i}) \array{\arrayopts{\align{center}} \begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="10em" height="10em" viewBox="0 0 120 120"> <desc>A_n Quiver</desc> <g fill="none" stroke="black" stroke-width="1"> <path d="M 60 25 l 25 20 l 0 30 l -25 20 l -25 -20 l 0 -30 z" /> <g stroke-dasharray="2"> <path d="M 60 0 l 0 25" /> <path d="M 85 45 l 25 -20" /> <path d="M 85 75 l 25 20" /> <path d="M 60 95 l 0 25" /> <path d="M 35 75 l -25 20" /> <path d="M 35 45 l -25 -20" /> </g> </g> <g fill="red" stroke="none"> <circle cx="60" cy="25" r="4" /> <circle cx="85" cy="45" r="4" /> <circle cx="85" cy="75" r="4" /> <circle cx="60" cy="95" r="4" /> <circle cx="35" cy="75" r="4" /> <circle cx="35" cy="45" r="4" /> </g> <foreignObject x="45" y="0" width="16" height="20"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mn>1</mn></msub></math></foreignObject> <foreignObject x="83" y="15" width="16" height="20"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mn>2</mn></msub></math></foreignObject> <foreignObject x="5" y="25" width="20" height="25"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mrow><msub><mi>n</mi><mn>1</mn></msub></mrow></msub></math></foreignObject> </svg> \end{svg} } \equiv \left({U(k)}^{n_1},\{v_i\}\right)

“This is my text”, says Anymouse.

Ruby code example:

    class Person
      attr_reader :name, :age
      def initialize(name, age)
        @name, @age = name, age
      end
      def <=>(person) # Comparison operator for sorting
        @age <=> person.age
      end
      def to_s
        "#@name (#@age)"
      end
    end
 
    group = [
      Person.new("Bob", 33), 
      Person.new("Chris", 16), 
      Person.new("Ash", 23) 
    ]
 
    puts group.sort.reverse

  1. You can also refer to it as (2). Chacun à son goût!.

  2. For more information, see Wikipedia.